Question 114 of 280

A $\sqrt{34}m$ long ladder weighing $10kg$ leans on a frictionless wall. Its feet rest on the floor $3m$ away from the wall as shown in the figure. If ${F}_{f}$ and ${F}_{w}$ are the reaction forces of the floor and the wall, then ratio of $\dfrac{{F}_{w}}{{F}_{f}}$ will be :
(Use $g=10m{s}^{-2}$.)

A$\dfrac{6}{\sqrt{110}}$B$\dfrac{3}{\sqrt{113}}$C$\dfrac{3}{\sqrt{109}}$D$\dfrac{2}{\sqrt{109}}$

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Answer: (C) $\dfrac{3}{\sqrt{109}}$

Applying force balance in vertical direction
${N}_{1}=mg=100N$
Applying torque balance at the bottom contact point,
$mg\times \dfrac{3}{2}={N}_{2}\times 5\Rightarrow {N}_{2}={F}_{w}=30N$
Applying force balance in horizontal direction
${f}_{s}={N}_{2}=30N$
Net reaction at the floor will be,
${F}_{f}=\sqrt{{{N}_{1}}^{2}+{{f}_{s}}^{2}}=\sqrt{{100}^{2}+{30}^{2}}=10\sqrt{109}N$
Therefore, $\dfrac{{F}_{w}}{{F}_{f}}=\dfrac{30}{10\sqrt{109}}=\dfrac{3}{\sqrt{109}}$